Recall that an endomorphism \(h:G\longrightarrow G\)  of a group \(G\)  is called an idempotent if \(h\circ h=h\) . It is clear that if \(H\)  is an \(r\) -image of \(G\)  with \(r\) -homomorphism \(g :G\longrightarrow H\)  and converse homomorphism \(f :H \longrightarrow G\) , then \(f\circ g:G\longrightarrow G\)  is an idempotent. By [1]}, \(C(G)\le e(G)\) , where \(e(G)\)  denotes the number of idempotent endomorphisms on \(G\) .

Let \(G\)  be a finite group of order \(n\) . Then \(C(G)\le n^{(\frac{1}{2}+o(1))\log _2 (n)}\) .


Clearly, \(e(G)\)  is equal to the number of pairs of subgroups \(H,K\)  such that \(G\)  is the semidirect product of \(H\)  and \(K\) . By [2]}, the number of subgroups of \(G\)  is bounded by \(n^{(\frac{1}{4}+o(1))\log _2 (n)}\) . Squaring gives an upper bound on the total number of pairs \((H,K)\)  and hence on \(e(G)\)  as follows:
\(e(G)\le n^{(\frac{1}{2}+o(1))\log _2 (n)}.\) 
